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A chain of mass M and length L is hold vertically onSmooth horizontal table as shown. If chain is released fromrest then find the normal reaction as a function ofx. x is the distance which the chain has fallen& assume norebounding of chain after striking the surface |
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Answer» Answer: Lets take a SMALL element at a distance x of length dx. Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor, Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is, Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx] Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdp Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the FORCE entered by length x due to its own WEIGHT is, Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=Lmgx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore, Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,Total force=Lmgx+L2mgx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,Total force=Lmgx+L2mgxF=L3mgx Lets take a small element at a distance x of length dx.The velocity with which element will strike the floor,v=2gxMomentum transferred to floor is,dP=dmv=[LMxdx2gx]F1=dtdpand, v=dtdx=2gxF1=LM2gx×2gxF1=L2mgxNow, the force entered by length x due to its own weight is,w=LmgxTherefore,Total force=Lmgx+L2mgxF=L3mgxExplanation: HOPE its help you |
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